Volume of 0.1 M `K_2Cr_2O_7` required to oxidize 35 ml of 0.5 M `FeSO_4` solution is

Options

A. 29.2 ml
B. 145 ml
C. 175 ml
D. 58.9 ml

Answer

To solve the problem of finding the volume of 0.1 M \( K_2Cr_2O_7 \) required to oxidize 35 ml of 0.5 M \( FeSO_4 \) solution, we can follow these steps:

### Step 1: Write the balanced redox reaction
The first step is to determine the balanced chemical equation for the reaction between \( K_2Cr_2O_7 \) and \( FeSO_4 \).

In this reaction, \( K_2Cr_2O_7 \) (potassium dichromate) is reduced from chromium in the +6 oxidation state to chromium in the +3 oxidation state, while \( FeSO_4 \) (iron(II) sulfate) is oxidized from iron in the +2 oxidation state to iron in the +3 oxidation state.

The balanced equation is:
\[ K_2Cr_2O_7 + 6FeSO_4 + 14H^+ \rightarrow 2Cr^{3+} + 6Fe^{3+} + 7H_2O + K^+ \]

### Step 2: Determine the n-factor for \( K_2Cr_2O_7 \)
The n-factor for a substance in a redox reaction is defined as the total number of electrons gained or lost per formula unit of the substance.

For \( K_2Cr_2O_7 \):
- Each chromium goes from +6 to +3, which is a change of 3.
- Since there are 2 chromium atoms, the total change is \( 2 \times 3 = 6 \).

Thus, the n-factor for \( K_2Cr_2O_7 \) is 6.

### Step 3: Determine the n-factor for \( FeSO_4 \)
For \( FeSO_4 \):
- Iron goes from +2 to +3, which is a change of 1.
- There is 1 iron atom, so the n-factor is 1.

### Step 4: Calculate the equivalents of \( FeSO_4 \)
To find the equivalents of \( FeSO_4 \), we use the formula:
\[ \text{Equivalents} = \text{Molarity} \times \text{Volume (L)} \times \text{n-factor} \]
Given:
- Molarity of \( FeSO_4 = 0.5 \, M \)
- Volume of \( FeSO_4 = 35 \, ml = 0.035 \, L \)
- n-factor of \( FeSO_4 = 1 \)

Calculating the equivalents:
\[ \text{Equivalents of } FeSO_4 = 0.5 \times 0.035 \times 1 = 0.0175 \, equivalents \]

### Step 5: Set up the equation for \( K_2Cr_2O_7 \)
Since the equivalents of \( K_2Cr_2O_7 \) must equal the equivalents of \( FeSO_4 \):
\[ \text{Equivalents of } K_2Cr_2O_7 = \text{Equivalents of } FeSO_4 \]
Let \( V \) be the volume of \( K_2Cr_2O_7 \) in liters. The equivalents can also be calculated as:
\[ \text{Equivalents of } K_2Cr_2O_7 = \text{Molarity} \times \text{Volume (L)} \times \text{n-factor} \]
Given:
- Molarity of \( K_2Cr_2O_7 = 0.1 \, M \)
- n-factor of \( K_2Cr_2O_7 = 6 \)

Setting the equations equal:
\[ 0.1 \times V \times 6 = 0.0175 \]

### Step 6: Solve for \( V \)
\[ 0.6V = 0.0175 \]
\[ V = \frac{0.0175}{0.6} = 0.0291667 \, L \]
Converting to milliliters:
\[ V = 0.0291667 \times 1000 = 29.17 \, ml \]

### Final Answer
The volume of 0.1 M \( K_2Cr_2O_7 \) required to oxidize 35 ml of 0.5 M \( FeSO_4 \) solution is approximately **29.17 ml**.

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